# Dirction of fastest growth

1. Nov 15, 2005

### sibiryk

I have f(x,y,z)=(x^2)y-x(e^z) and point Po=(2,-1,pi)
I need to find
a) gradient at point Po ( done)
b) Rate of change of f at point Po in the direction of vector u=i-2j+k (it's also done)
c) Unit vector in the direction of fastest growth of f at Po.

I can't find formulas for a last on. Does it come from a) and b)?
I know that angle should be zero but I am not sure what angle it is.

2. Nov 15, 2005

### whozum

At what angle is the growth greatest? Its got to do with the directional derivative and the gradient.

3. Nov 15, 2005

### sibiryk

The growth is greatest when angle equals zero.

4. Nov 15, 2005

### HallsofIvy

Staff Emeritus
When WHAT angle equals zero?

(You are almost there!)

5. Nov 15, 2005

### sibiryk

I think it's the angle between unit vector and tangent vector to f at Po.
If I'm correct, I'm not sure how should I put it on a paper.

6. Nov 15, 2005

### whozum

Youre right. When you take a directional derivative what kind of product do you use (for the two vectors). One of these two is at a maximum when the angle is zero, thats the one you want to use.

If you're still stuck, show the defining expression for a directional derivative.

Last edited: Nov 15, 2005
7. Nov 15, 2005

### HallsofIvy

Staff Emeritus
Not "tangent vector" because you are not talking about a curve.

The gradient is the "derivative" of a function of 2 or more variables. The derivative of f(x,y) in the direction of angle $\theta$ if given by
$$D_\theta f(x,y)= (cos \theta i+ sin \theta j)\dot(\frac{\partial f}{\partial x}i+ \frac{\partial f}{\partial y})$$
That will have a maximum (with respect to $\theta$) where it's derivative with respect to $\theta$ equals 0:
$$-cos\theta \frac{\partial f}{\partial x}+ sin\theta\frac{\partial f}{\partial y}= 0$$
That means that
$$tan\theta= \frac{sin\theta}{cos\theta}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$

Think about what that means in terms of the components of the gradient of f.

8. Nov 15, 2005

### whozum

It is tangent to the surface at that point and I think that is what he is meaning, but when youre dotting $D_{\theta}f(x,y)$ why did we get a negative cosine?

You also forgot your j unit vector in the first expression.

9. Nov 15, 2005

### sibiryk

Yes, my mistake I thought of tangent vector to a surface.
Why should I dot f(x,y) if I have f(x,y,z)?
Still confused.